Problem: As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as <image>, where k is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.

A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number n, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!

Input

The first input line contains an integer n (1 ≤ n ≤ 109).

Output

Print "YES" (without the quotes), if n can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).

Examples

Input

256


Output

YES


Input

512


Output

NO

Note

In the first sample number <image>.

In the second sample number 512 can not be represented as a sum of two triangular numbers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline T Max(T a, T b) {
  if (a > b)
    return a;
  else
    return b;
}
template <class T>
inline T Min(T a, T b) {
  if (a < b)
    return a;
  else
    return b;
}
template <class T>
inline T gcd(T a, T b) {
  if (a < 0) return gcd(-a, b);
  if (b < 0) return gcd(a, -b);
  return (b == 0) ? a : gcd(b, a % b);
}
template <class T>
inline T lcm(T a, T b) {
  if (a < 0) return lcm(-a, b);
  if (b < 0) return lcm(a, -b);
  return a * (b / gcd(a, b));
}
template <class T>
inline T TripleMax(T a, T b, T c) {
  return Max(Max(a, b), c);
}
template <class T>
inline T TripleMin(T a, T b, T c) {
  return Min(Min(a, b), c);
}
long long n;
long long s;
int solve(long long data) {
  long long ret;
  ret = (long long)sqrt(double(2 * data));
  if (ret * (ret + 1) / 2 == data || ret * (ret - 1) / 2 == data) {
    return 1;
  } else {
    return 0;
  }
}
int main(int argc, const char* argv[]) {
  cin >> n;
  s = (long long)sqrt(double(2 * n));
  for (int i = 1; i < s + 1; i++) {
    if (solve(n - i * (i + 1) / 2) == 1 && n - i * (i + 1) / 2 != 0) {
      cout << "YES" << endl;
      return 0;
    }
  }
  cout << "NO" << endl;
  return 0;
}